Binomial Probability Riddle

[uva3021]

50 Point reward for best response. I'm not sure what the answer is, I have an idea how to do it. Its a series of cumulative binomial probability calculations. But I have yet to arrive at an answer I would assert as being correct with a high confidence level.

[Kramer]

Does this help ?



Cumulative Binomial Probability
A cumulative binomial probability refers to the probability that the binomial random
variable falls within a specified range

(e.g., is greater than or equal to a stated lower limit and less than or equal to
a stated upper limit).


For example, we might be interested in the cumulative binomial probability
of obtaining 45 or fewer heads in 100 tosses of a coin
(see Example 1 below).

This would be the sum of all these individual binomial probabilities.

b(x < 45; 100, 0.5) =
b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + ...
+ b(x = 44; 100, 0.5) + b(x = 45; 100, 0.5)


CALCULATOR PASTED BELOW ...

AP* Statistics Tutorial: Binomial Distribution

[uva3021]

lol just checked this wtf my image got deleted, or perhaps it was never there

surely someone would have said something :confused:

[Kramer]

I think I got it -

did you flip the 10 coins all at once ,
or independantly of each other ?

When we toss a coin, there are two possibly outcomes.

It can be a head or a tail, which are both equally likely.

If we toss two coins, there can be two heads, two tails, or a head and a tail.

It is tempting to say that there are three equally possible outcomes.
But this would be wrong.
You must think of the coins separately.

It might be easier to imagine tossing one coin first and the other after
(or even tossing the same coin twice, which has exactly the same effect).


Or you could imagine two different values of coins, so they can be told apart.
Now you can see that there are four possibilities: both heads, both tails,
first coin a head and the second a tail, and first coin a tail and the second a head


Say that you are going to toss three coins, and you want to work out
the probability of only one head (and so two tails).

The possible outcomes are:


TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
All these outcomes are different, and they are all equally likely.

There are 8 of them. There are 3 tosses with only one head:
TTH, THT, HTT
So the probability is 3/8.

You can convert this into a decimal 0.375 or a percentage 37.5%,

which you can round to 38% if you wish.
Or you can describe it as a three in eight chance.
All these mean the same.

You can list the possible outcomes above for any dice up to 6

and count the tosses which match the probability that you want.

With 10 coins, there are 1024 possible outcomes




Probability - theory of tossing coins

[thom321]

I know it is a bit old but since I just saw it, I decided to have my crack at it.
First off, I am not a mathematician or statistician so I would bet a very small number of units on my answer being perfectly correct. However, here is my shot at it.

As far as I understand it, we are dealing with both conditional and independent probabilities.

Tossing 10 coins has a 2^10 = 1024 possible outcomes, of which 968 has at least 3 heads = probability of 968/1024 = 0.9453.

As an independent event, tossing 5 coins and getting at least 2 heads has 32 possible outcomes of which 26 has at least 2 heads = probability of 26/32 = 0.8125.

From my understanding, we are randomly selecting 5 coins, after 10 coins have been tossed that had at least 3 heads. Let’s call tossing the 10 coins and getting 3 heads “Event A” and selecting 5 coins and getting at least 2 heads “Event B”. So we are looking at the probability of “Event B” given that “Event A” has already happened. The probability of B given A is the same as (independent probability of B x independent probability of A) divided by independent probability of A. In this case it would mean the following: Probability of getting at least 2 heads when randomly picking 5 coins out of the previously tossed 10 where we had at least 3 heads = (probability of getting at least 3 heads out of 10 tossed coins x probability of getting at least 2 heads out of 5 tossed coins)/ probability of getting at least 3 heads out of 10 tossed coins = probability of getting at least 2 heads out of 5 tossed coins = 0.8125 i.e. the same as the independent probability for tossing 5 coins and getting at least 2 heads.

So as long as the result of initial coin toss does not come with conditions that are mutually exclusive to the second random selection of coins from the initial toss, the probability of the random selection is the same as if you had tossed the coins as an independent event.

I verified this by setting up an example with 4 “coins” and then “selecting” 3. When tossing 4 coins there are 2^4 = 16 possible outcomes. From each 4 coins, you can create 4 combinations of 3 coins (i.e. you select 3 coins from the 4 tossed), which gives us 64 combinations of 3 coins. If we require the first toss of 4 coins to have at least 1 head, we eliminate 1 (there is only one way to have less than 1 head i.e. all tails) of the 16 combinations, leaving us with 15. So the probability of getting at least 1 head when tossing 4 coins is 15/16 = 0.9375. 15 combinations of 4 coins = 60 possible combinations of 3 coins. If we then randomly select 3 coins and require at least 2 heads, out of the 60 remaining possible combinations, 32 of them will have at least 2 heads. 32/60 = 0.533333. So the combined probability of those two events is 0.9375 x 0.53333 = 0.5. Compare this to tossing 3 coins and getting at least 2 heads. When tossing 3 coins, there are 8 possible combinations. 4 of those have at least 2 heads, making the probability 4/8 = 0.5, which is the same probability that I got above when I first “tossed” the 4 coins and requiring at least 1 head and then selecting 3 coins and requiring at least 2 heads.

This in turn means that in our example, we can focus on the last part of the question. In order for the friend to be left with two heads, one of three things must have happened

1. The selection of 5 coins must have had 3 heads AND the selection of 3 coins must have had 1 head

OR

2. The selection of 5 coins must have had 4 heads AND the selection of 3 coins must have had 2 heads

OR
3. The selection of 5 coins must have had 5 heads AND the selection of 3 coins must have had 3 heads

The probability of getting 3 heads when tossing 5 coins = 10 matching outcomes from a total number of possible combinations of 32 = 10/32 = 0.3125
The probability of getting 1 head when tossing 3 coins = 3 matching outcomes from a total number of possible combinations of 8 = 3/8 = 0.375. The joint probability of these two events is 0.3125 x 0.375 = 0.117188

The probability of getting 4 heads when tossing 5 coins = 5 matching outcomes from a total number of possible combinations of 32 = 5/32 = 0.15625
The probability of getting 2 heads when tossing 3 coins = 3 matching outcomes from a total number of possible combinations of 8 = 3/8 = 0.375. The joint probability of these two events is 0.15625 x 0.375 = 0.058594

The probability of getting 5 heads when tossing 5 coins = 1 matching outcome from a total number of possible combinations of 32 = 1/32 = 0.03125
The probability of getting 3 heads when tossing 3 coins = 1 matching outcome from a total number of possible combinations of 8 = 1/8 = 0.125. The joint probability of these two events is 0.03125 x 0.125 = 0.003906.

Since either of these would lead to the desired outcome, we can add up the probabilities of each to get to our total probability. 0.117188 + 0.058594 + 0.003906 = 0.179688 = 17.97%. That is my answer.

FYI, I used Probability - theory of tossing coins to come up with the number of matching combinations for tossing 10 and 5 coins but I wrote down the examples with 3 and 4 “coins” in Excel and listed all combinations myself.

[uva3021]

Great post thom

The confusion in this problem arises from the wording. While intuitively one would think you would apply conditional probability theory, the use of the word "if" is not included here and the events have already happened. Thus all the probabilities are implied in the independent binomial distribution calculations. These just appear to happen in sequence. Using this reasoning, you calculate each event independently then multiply the probabilities.

The answer I get is 16.80%.

You may be right, I don't know, this is how I reasoned it. I've found a few on the internet who agree with my assessment, and have found others who have approached the problem similar to how you approached it.